Question

A vehicle is moving with a velocity v on a curved road of width b and radius of curvature R. For counteracting the centrifugal force on the vehicle, the difference in elevation required between the outer and inner edges of the road is?

• A. $\frac{\mathrm{bv}}{\mathrm{Rg}}$
• B. $\frac{{\mathrm{vb}}^2}{\mathrm{Rg}}$
• C. $\frac{{\mathrm{bv}}^2}{\mathrm{Rg}}$
• D. $\frac{{\mathrm{Rg}}^2}{\mathrm{bv}}$

Answer 1

The correct option is C

$\frac{{\mathrm{bv}}^2}{\mathrm{Rg}}$

Step 1: Given information

The vehicle moves with the velocity “v”.

The radius of curvature is “R”.

Width “b”.

Step 2: To find

We have to find the difference in elevation required between the outer and inner edges of the road.

Step 3: Calculation

For the vehicle banking on a curved path,

$\mathrm{tan\theta }=\frac{{\mathrm{v}}^2}{\mathrm{rg}}...\left(1\right)$

By using the reference figure,

Here,

$\mathrm{\theta }$ is small

$\mathrm{sin\theta }=\frac{\mathrm{h}}{\mathrm{b}}=\mathrm{tan\theta }...\left(2\right)$

By equating the equations $\left(1\right)$ and $\left(2\right)$, we get

$$\begin{gathered} \Rightarrow \frac{\mathrm{h}}{\mathrm{b}}=\frac{{\mathrm{v}}^2}{\mathrm{Rg}} \\ \mathrm{h}=\frac{{\mathrm{bv}}^2}{\mathrm{Rg}} \end{gathered}$$

Therefore, the difference in elevation required between the outer and inner edges of the road is $\frac{{\mathrm{bv}}^2}{\mathrm{Rg}}$.